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Use $a_n=N^{-1/2}{\sum }_k{e}^{ikn}{a}_k$ with $k=2\pi j/N$, choosing $N$ distinct values modulo $2\pi$ in the Brillouin zone. The lattice spacing is one; for spacing $a$ the dimensionless variable here would be $ka$. Orthogonality of the discrete Fourier modes diagonalizes the bilinear Hamiltonian:This is the nearest-neighbour ferromagnetic magnon dispersion. The printed expression is missing a factor of four for the stated exchange Hamiltonian. If ${\omega }_k$ denotes a physical frequency, $\hslash {\omega }_k={\epsilon }_k$; setting $\hslash =1$ does not remove that factor. The printed coefficient would require an exchange constant $J/4$ in the original Hamiltonian.

An independent check uses one spin lowering at site $n$. The exact one-magnon Hamiltonian relative to $E_0$ has diagonal element $2JS$ and off-diagonal elements $-JS$ at $n\pm 1$, giving the same Fourier eigenvalue. At small $k$, ${\epsilon }_k\simeq JS{k}^2$, and the $k=0$ state belongs to the degenerate ground multiplet. The quadratic low-energy branch is the ferromagnetic type-B Goldstone boson.

Apply a discrete Fourier transform, with lattice spacing one:The wavevectors lie in a Brillouin zone, which we choose as $[-\pi ,\pi )$. Pairing $k$ with $-k$ makes the quadratic HamiltonianA bosonic Bogoliubov diagonalization uses $a_k=\cosh {\theta }_k{\alpha }_k-\sinh {\theta }_k{\alpha }_{-k}^{†}$. The canonical commutation relations hold because ${\cosh }^2{\theta }_k-{\sinh }^2{\theta }_k=1$. The anomalous terms vanish when $\tanh 2{\theta }_k=B_k/A=\cos k$, givingNear $k=0$ and $k=\pi$, the dispersion relation is linear: ${\omega }_k\sim 2JS|k-k_0|$. These are the low-energy antiferromagnetic spin waves; the two zeroes are related by the two-sublattice description. The transformation is singular at the exact zero modes, which require an infrared regulator or separate treatment of the collective rotation.

The normalized Bloch states $|k⟩=N^{-1/2}{\sum }_{n=1}^N{e}^{ikna}|n⟩$ satisfy periodicity when $k=2\pi j/(Na)$. Acting with the hopping terms shifts the coefficients by one site, givingThe first Brillouin zone is a fundamental cell of reciprocal space, for example $[-\pi /a,\pi /a)$; wavevectors differing by $2\pi /a$ label the same state. It contains exactly $N$ distinct allowed wavevectors and hence $N$ orbital states. The written Hamiltonian has no spin index; if electron spin degeneracy is included there are $2N$ states.

For a narrow wave packet, group velocity and band-curvature effective mass areFor the usual $J>0$, the mass is negative in the outer half of the zone, $\pi /(2a)<|k|\le \pi /a$, taking one endpoint representative. At $|k|=\pi /(2a)$ the inverse mass vanishes and the mass diverges. For arbitrary nonzero $J$, the sign criterion is $J\cos ka<0$; $J=0$ is a flat band.

For electron charge $-e$ with $e>0$ in a constant electric field $\mathcal{E}$, the single-band semiclassical equations are $\hslash \dot{k}=-e\mathcal{E}$, $\dot{x}=v(k)$. Thus $k(t)=k_0-e\mathcal{E}t/\hslash$, and for $\mathcal{E}\ne 0$,These are Bloch oscillations of period $2\pi \hslash /(|e\mathcal{E}|a)$. The zero-field limit is $x-x_0=v(k_0)t$. This assumes a narrow packet and neglects scattering and interband transitions.

For many electrons obeying the Pauli exclusion principle, a partially filled band has available nearby states and can carry current under a displaced occupation distribution: it describes a metal within band theory. A completely filled band has cancelling velocities over the zone, so its current is zero; if separated from empty higher bands by a nonzero gap, it describes an insulator at sufficiently low temperature and weak electric field. With spin degeneracy this isolated band is filled at two electrons per site. The single-electron Hamiltonian alone does not specify the filling or a gap to a next band, so those extra physical assumptions are required for classifying a material.

Label the two atoms in primitive cell $j$ by $u_{2j}$ and $u_{2j+1}$, with $j$ understood modulo $N$ because of the periodic boundary condition. Choose the spring between them to have constant $\lambda$. Newton's second law gives

For a longitudinal normal mode, putPeriodicity requires $e^{2iNka}=1$, so $k=\pi r/(Na)$. Since the primitive-cell length is $2a$, wavevectors differing by $\pi /a$ describe the same mode, and the first Brillouin zone may be chosen asThe amplitudes obeySetting its determinant to zero gives the alternating-spring chain dispersion relationThe lower sign is the acoustic phonon branch and the upper sign is the optical phonon branch.

At the edge of the Brillouin zone,so the frequency band gap isIt vanishes at $\alpha =1$, when the apparent two-atom primitive cell can be reduced to one atom.

Near the centre of the Brillouin zone, a Taylor expansion givesandThe group velocity of the acoustic branch at long wavelength is the speed of sound,

A three-dimensional Bravais lattice is the set of all integer combinations of three linearly independent primitive vectors. Its reciprocal lattice isand the first Brillouin zone is the Wigner-Seitz cell of the origin in reciprocal space.

For the stated FCC lattice, the primitive-cell volume isThe reciprocal-basis formula givesA reciprocal primitive cell, and hence the Brillouin zone, has volume

In the first Born approximation, with momentum transfer $q=k-k^{\prime }$, the scattering amplitude isFor the spherical shell, align the polar axis with $q$ and use its radial Dirac delta function:Therefore

Translation of each atomic potential multiplies its transform by a phase, so the crystal lattice structure factor givesFor an infinite lattice, the lattice Poisson summation formula turns the sum intoThus the amplitude vanishes away from reciprocal-lattice momentum transfers.

In elastic scattering, $|k|=|k^{\prime }|$ and henceThe two shortest relevant reciprocal-vector lengths of this FCC lattice areBoth are kinematically available when $k>2\pi /a$. Assigning their scattering angles ${\theta }_2$ and ${\theta }_1$, respectively, gives

Bloch theorem states that the energy eigenstates of a Hamiltonian invariant under translations by a Bravais lattice can be chosen as Bloch statesIndeed, the unitary operators $T_r$ commute with one another because $T_r{T}_s=T_{r+s}=T_s{T}_r$, and they commute with $H$ by hypothesis. The simultaneous diagonalization theorem therefore lets us diagonalize all translations within each energy eigenspace. Their eigenvalues form a unitary character $\chi$ of the additive lattice:Writing $\chi (a_j)=e^{-ik\cdot a_j}$ on a primitive basis gives $\chi (r)=e^{-ik\cdot r}$. In position space, with $(T_r\psi )(x)=\psi (x-r)$, this implies ${\psi }_k(x+r)=e^{ik\cdot r}{\psi }_k(x)$. Hence $u_k(x)=e^{-ik\cdot x}{\psi }_k(x)$ is lattice-periodic. Adding a reciprocal lattice vector to $k$ leaves the character unchanged, so the crystal momentum lies in a Brillouin zone.

Forthe equations $a_i\cdot b_j=2\pi {\delta }_{ij}$ give the reciprocal basisThe reciprocal lattice is triangular, so the first Brillouin zone is its Wigner-Seitz cell, a regular hexagon. Its six corners areReciprocal-lattice translations identify these corners in two classes of three. Representatives areFor example, $(b_1-b_2)/3=K^{\prime }-b_2$, while the other equivalences follow by symmetry and reciprocal translations. This gives the requested sketch: a regular hexagon with the vertical edge from $K^{\prime }$ to $K$ at $k_x=2\pi /(\sqrt{3}a)$ and alternating $K,K^{\prime }$ corner classes.

For the tight-binding model, introduce the normalized Bloch sumEach hop by $\pm a_j$ multiplies this state by $e^{\mp ik\cdot a_j}$. Thus the two-direction nearest-neighbour tight-binding dispersion isAlong the boundary edge from $K^{\prime }$ to $K$, $k_x=2\pi /(\sqrt{3}a)$ and $|k_y|\le 2\pi /(3a)$, soFor $\lambda >0$, it rises from $E_0+2\lambda$ at either corner to $E_0+4\lambda$ at the midpoint; for $\lambda <0$ the ordering reverses. Globally the two cosines in the first expression can simultaneously equal $1$ or $-1$, and therefore

Each orbital band contains two one-electron states per lattice site because an electron has two spin states. By band filling, a valency of one leaves this band half-filled and the material conducts, whereas a valency of two fills it. A filled band can be insulating only if it is separated from every empty band by a positive band gap.

The first band's maximum is $E_0+4|\lambda |$. If the second band's minimum is $E_0+\Delta$, a gap exists precisely when $\Delta >4|\lambda |$. Consequently, among the nontrivial fillings described here,For valency one the first band is partially filled. If $\Delta \le 4|\lambda |$, the bands overlap or touch, so even at valency two the overlapping energy bands prevent a band insulator.

For a lattice of period $a$, wavevectors differing by the reciprocal-lattice vector $2\pi /a$ describe the same translation eigenvalue. The first Brillouin zone may therefore be chosen as

Bloch theorem states that every energy eigenstate in a periodic potential may be chosen in the formTo prove it, let $T_a$ be translation by one period:Periodicity gives $[T_a,H]=0$. The translation operator is unitary, so it can be diagonalized within each energy eigenspace. Choose a common eigenstate,Unitarity gives $|\lambda |=1$, so $\lambda =e^{ika}$ for some real $k$. Thus $\psi (x+a)=e^{ika}\psi (x)$. Defining $u_k(x)=e^{-ikx}\psi (x)$ giveswhich proves the theorem.

A Bravais lattice is the integer span of three linearly independent primitive vectors. Its reciprocal lattice consists of vectors $G$ satisfying $G\cdot r\in 2\pi \mathbb{Z}$ for every lattice vector $r$.

Translations by lattice vectors commute with the Hamiltonian. Their simultaneous eigenvalues have the form $e^{ik\cdot r}$, so an energy eigenfunction may be chosen withThus ${\psi }_k=e^{ik\cdot x}{u}_k$ with lattice-periodic $u_k$, which is Bloch theorem. Wavevectors differing by a reciprocal-lattice vector describe the same translation character, and the Brillouin zone is a fundamental Wigner-Seitz cell of the reciprocal lattice.

Every primitive basis of the Bravais lattice has the formwhere $p,q,r,s\in \mathbb{Z}$ andThe unimodular determinant condition is exactly what makes the integer change of basis invertible over $\mathbb{Z}$.

Let $b_i$ be the primitive vectors of the reciprocal lattice, withSolving these four equations givesThe six reciprocal-lattice points nearest the origin areall at distance $4\pi /\sqrt{3}$.

The Wigner-Seitz cell is bounded by the perpendicular bisectors of the segments joining the origin to those six points. It is the regular hexagon with verticesIts area equals the reciprocal primitive-cell area:This is the reciprocal lattice and Wigner-Seitz cell of the unit triangular lattice.

For a periodic potential, Bloch theorem identifies wavevectors differing by a reciprocal-lattice vector. The first Brillouin zone is the Wigner--Seitz cell just found. More generally, the $n$th Brillouin zone consists of wavevectors reached from the origin after crossing exactly $n-1$ reciprocal-lattice Bragg planes; its boundaries are the perpendicular bisectorsFor this triangular reciprocal lattice, the first zone is the central regular hexagon. The second is the sixfold-symmetric collection of regions immediately outside its six sides, bounded next by the bisectors associated with the next reciprocal points. These zones organize free-particle states into bands: Bragg coupling is strongest at their boundaries and opens energy gaps when the periodic potential is introduced.

Translation by one site multiplies a lattice wave by $e^{ika}$, so the coefficients of a finite periodic tight-binding ring arePeriodicity after three sites requiresChoosing representatives in the Brillouin zonegivesThe corresponding states areThey reproduce respectively the coefficient vectors $(1,1,1)$, $(1,\omega ,{\omega }^2)$ and $(1,{\omega }^2,\omega )$.